Fundamental Solution of the Laplacian In All Dimensions
The Laplacian on $\mathbb{R}^n$ is the partial differential operator defined as $$\nabla=\sum_{j=1}^n\frac{\partial^2}{\partial x_j^2}$$
In this article, we wish to find fundamental solutions to this operator. That is, we seek a distribution $F\in\mathcal{D}(\mathbb{R}^n)$ such that $\nabla F$ is the Dirac distribution. Why? Suppose we want to find a function $f$ such that $\nabla f=\phi$ where $\phi$ is a compactly supported smooth function.
If we set $f=F*\phi$ (where $F$ is defined respectively for the two different cases), then $\Delta f=(\Delta F)*\phi=\delta*\phi$ using Theorem 6.30 (b) (Grandpa Rudin) and the fact that $\Delta$ is a linear differential operator. Now, $(\delta*\phi)(x)=\langle\delta,\tau_x\tilde{\phi}\rangle=\phi(x)$ for all $x\in\mathbb{R}^n.$ Hence, $\delta*\phi=\phi.$ So $\Delta f=\phi.$
We will divide our arguments into three cases: $n=1,$ $n=2,$ and $n\ge 3.$ The proof of Case II and Case III are almost identical (same steps, just different functions). Case II (where $n=2$) is the trickier of the two since it involves some annoying manipulation with log functions.
Motivation: I was motivated to write this article after coming out of a real analysis lecture in early April. Our professor was teaching us distributions and he went on a tangent to explain the concept the fundamental solutions and showed (in just 15 mins) how to find the solution of the Laplacian using Green’s theorem and divergence operators. Everyone in class understood it except the two undergrads in class, one being me. I had two problems:
(1) It was 9 o’ clock in the morning on a Friday
(2) I didn’t (and still don’t) know Green’s theorem rigorously
Hence, I present the longer but easier proof without using Greens theorem.
Case I: $n=1$
The function $F:x\mapsto\frac{|x|}{2}$ is obviously a locally integrable function on $\mathbb{R}.$ We claim that this (distribution) is a fundamental solution to the Laplacian in $\mathbb{R}.$
For any $\phi\in\mathcal{D}(\mathbb{R}),$ we have, by definition, that $$\langle \Delta F,\phi\rangle=\langle F,\Delta\phi\rangle=\int\frac{|x|}{2}\phi''(x)dx$$
Firstly, using dominated convergence theorem (as $x\mapsto \chi_{(0,\infty)}(x)\frac{x}{2}\phi''(x)$) is integrable $\big($since the function is non-zero only inside some compact set (which obviously has finite Lebesgue measure) and $\phi''$ is a bounded function$\big)$ and integration by parts (since the concerned integrands are atleast $\mathcal{C}^1$) $$\int_0^\infty\frac{x}{2}\phi''(x)dx=\lim_{M\to\infty}\int_0^M\frac{x}{2}\phi''(x)dx$$$$=\lim_{M\to\infty}\left[\frac{M}{2}\phi'(M)-\int_{0}^M\frac{\phi'(x)}{2}dx\right]=-\lim_{M\to\infty}\left[\frac{\phi(M)-\phi(0)}{2}\right]=\frac{\phi(0)}{2}$$
In the last line, we have used the fact that both $\lim_{M\to\infty}\left[\frac{M}{2}\phi'(M)\right]=0$ and $\lim_{M\to\infty}\left[\phi(M)\right]=0$ for the simple reason that $\phi$ (and, hence $\phi'$) has compact support (so $\phi(M)=\phi'(M)=0$ for all $M\ge M_0$ for some $M_0\in\mathbb{R}$). We have also used the fundamental theorem of calculus to get $\int_0^M\phi'(x)dx=\phi(M)-\phi(0).$
Using similar steps, we have $\int_{-\infty}^0\frac{-x}{2}\phi''(x)dx=\frac{\phi(0)}{2}.$ Hence, $$\langle \Delta F,\phi\rangle=\int_{-\infty}^0\frac{-x}{2}\phi''(x)dx+\int_0^\infty\frac{x}{2}\phi''(x)dx=\phi(0)=\langle \delta,\phi\rangle$$
Since $\phi\in\mathcal{D}(\mathbb{R})$ is arbitrary, we have our claimed fundamental solution.
Case II: $n=2$
Define functions $F$ on $\mathbb{R}^2$ as $F:x\mapsto\frac{1}{2\pi}\log\big(|x|)$ and $F^t$ on $\mathbb{R}^2$ for all $t>0$ as $F^t:x\mapsto\frac{1}{4\pi}\log\big(|x|^2+t^2\big).$ This is a locally integrable function since for any compact $K\subset \mathbb{R}^2,$ there exists some $R>0$ such that $K\subset B(0,R)$ (by Heine Borel theorem), so using polar co-ordinates and substituting $u=r^2+t^2$, $$\int_K \left|\frac{1}{4\pi}\log\big(|x|^2+t^2\big)\right|dx\le \int_{\mathcal{S}^1}\int_0^R\frac{1}{4\pi}r|\log(r^2+t^2)|drd\sigma(x')$$$$=\int_{\mathcal{S}^1}\int_{t^2}^{R^2+t^2}\frac{|\log(u)|}{8\pi}dud\sigma(x')=\frac{\sigma(\mathcal{S}^1)}{8\pi}\int_{t^2}^{R^2+t^2}|\log(u)|du$$
Now, $$\int_{t^2}^{R^2+t^2}|\log(u)|du=\int_{t^2}^{1}-\log(u)du+\int_{1}^{R^2+t^2}\log(u)du$$$$=\left[(1+t^2\log(t^2)-t^2)+\big((R^2+t^2)\log(R^2+t^2)-(R^2+t^2)+1\big)\right]<\infty$$
Since $\sigma(\mathcal{S}^1)<\infty,$ our claim is proved. Further, the function $F$ is also $L^1_\text{loc}$ since (using the same notation and steps as before), $$\int_K \left|\frac{1}{2\pi}\log\big(|x|\big)\right|dx\le \frac{\sigma(\mathcal{S}^1)}{2\pi}\int_0^Rr|\log(r)|dr$$
Again, $$\int_0^Rr|\log(r)|dr=\int_0^1-r\log(r)dr+\int_1^Rr\log(r)dr$$
The second integral is finite because it is equal to $\frac{R^2\log R}{2}-\frac{R^2}{4}+\frac{1}{4}$ using integration by parts. The first can be computed by monotone convergence theorem as $\lim_{M\to 0^+}\int_M^1-r\log rdr=\frac{1}{4}$ since $\lim_{M\to 0^+}M^2\log(M)=0.$ Thus, $F$ is indeed locally integrable (as $\sigma(\mathcal{S}^1)<\infty$).
Now consider any $\phi\in\mathcal{D}(\mathbb{R}^2).$ Say $\text{supp}(\phi)\subset [-M/2,M/2]^2$ for some $M>0$ (since the support is compact and, hence, bounded by Heine Borel theorem). Then, for any $t>0,$ we have, using Fubini's theorem (as the integrand is $L^1$), $$\left\langle F^t,\frac{\partial^2}{\partial x^2}\phi\right\rangle=\int F^t(x,y)\left(\frac{\partial^2}{\partial x^2}\phi\right)(x,y)d(x,y)=\int_{-M}^M\frac{1}{4\pi}\int_{-M}^M\log(x^2+y^2+t^2)\left(\frac{\partial^2}{\partial x^2}\phi\right)(x,y)dx dy$$
For any $y\in\mathbb{R},$ we can simplify the inner integral using integration by parts (which is justified as integrands are atleast $\mathcal{C}^1$ because $t>0$): $$\int_{-M}^M\log(x^2+y^2+t^2)\left(\frac{\partial^2}{\partial x^2}\phi\right)(x,y)dx=-\int_{-M}^M\frac{2x}{x^2+y^2+t^2}\left(\frac{\partial}{\partial x}\phi\right)(x,y)dx$$
This is because $\left(\frac{\partial}{\partial x}\phi\right)(\pm M,y)=0$ by definition of $M.$ Again, as $t>0,$ the integrands are atleast $\mathcal{C}^1,$ so this is equal to $$\int_{-M}^M\frac{2y^2-2x^2+2t^2}{\big(x^2+y^2+t^2\big)^2}\phi(x,y)dx\text{ }\Big(\text{as $\phi(\pm M,y)=0$}\Big)$$
We now compute $\left\langle F^t,\frac{\partial^2}{\partial y^2}\phi\right\rangle$ similarly with a flipped order of integration (valid by Fubini's theorem). Then, adding them together, and reapplying Fubini's theorem, to convert the double integral back to an integral in $\mathbb{R}^2,$ we have $$\int F^t(x,y)\Bigg[\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\phi\Bigg](x,y)d(x,y)=\frac{1}{4\pi}\int_{[-M,M]^2}\frac{4t^2}{\big(x^2+y^2+t^2\big)^2}\phi(x,y)d(x,y)\text{ }(\ddagger)$$
Define function $g$ on $\mathbb{R}^2$ as $g:(x,y)\mapsto\frac{1}{\pi}\frac{1}{\big(x^2+y^2+1\big)^2}$ and $g^t$ on $\mathbb{R}^2$ for all $t>0$ as $g:(x,y)\mapsto t^{-2}g(t^{-1}(x,y)).$ Since $\phi\in\mathcal{D}(\mathbb{R}^2)$ is arbitrary, one can verify from $(\ddagger)$ that $\Delta F^t=g_t$ distributionally.
We note that $g\in L^1(\mathbb{R}^2)$ since (using polar coordinates and substituting $u=r^2+1$) $$\int |g|dm=\int gdm=\frac{1}{\pi}\int_{\mathcal{S}^1}\int_0^\infty\frac{r}{(r^2+1)^2}drd\sigma(x')=\frac{\sigma(\mathcal{S}^1)}{2\pi}\int_1^\infty\frac{1}{u^2}du=\frac{\sigma(\mathcal{S}^1)}{2\pi}=1<\infty$$
Using Proposition 9.1 (Folland), $\Delta F^t=g_t\longrightarrow\delta$ in $\mathcal{D}'(\mathbb{R}^2)$ as $t\longrightarrow 0.$ ($\star$)
On the other hand, for any $0<t<1$ and $z\in\mathbb{R}^2$ such that $z\neq 0$ note that:
Case (i): If $0<|z|^2+t^2\le 1$ then $0<|z|^2<|z|^2+t^2\le 1$ and using monotonicity of the $\log$ function, we have $2\log|z|<\log(|z|^2+t^2)\le 0.$ Hence, $$\big|\log(|z|^2+t^2)\big|< 2\big|\log(|z|)\big|\le 2\big|\log(|z|)\big|+\big|\log(|z|^2+1^2)\big|$$
Case (ii): If $|z|^2+t^2>1$ then $1<|z|^2+t^2<|z|^2+1^2$ and using monotonicity of the $\log$ function, we have $0<\log(|z|^2+t^2)<\log(|z|^2+1^2).$ Hence, $$\big|\log(|z|^2+t^2)\big|<\big|\log(|z|^2+1^2)\big|\le 2\big|\log(|z|)\big|+\big|\log(|z|^2+1^2)\big|$$
Thus, we conclude that $|F^t|\le |F^1|+|F|$ a.e when $0<t<1$ (actually everywhere except the origin).
Consider any sequence $(t_m)_{m=1}^\infty\subset (0,1)$ which converges to $0$ as $m$ approaches infinity. For any $\phi\in\mathcal{D}(\mathbb{R}^2),$ we thereby have $$\lim_{m\to\infty}\langle F^{t_m},\phi\rangle=\lim_{m\to\infty}\int \frac{\log(|z|^2+t_m^2)}{4\pi}\phi(z)dz=\int \lim_{m\to\infty}\frac{\log(|z|^2+t_m^2)}{4\pi}\phi(z)dz=\int \frac{\log(|z|)}{2\pi}\phi(z)dz=\langle F,\phi\rangle$$
This follows from the dominated convergence theorem since both $F^1,F\in L^1_\text{loc}(\mathbb{R}^2)$ as shown before and $|F^{t_m}\phi|\le |F^1\phi|+|F\phi|$ a.e (where $|F^1\phi|+|F\phi|\in L^1(\mathbb{R}^2)$ since $\phi$ has compact support and is bounded) for all $m\in\mathbb{N}.$
Since $\phi$ is arbitrary, we have $F^t\longrightarrow F$ (hence, also $\nabla F^t\longrightarrow \nabla F$) in $\mathcal{D}'(\mathbb{R}^2)$ as $t\longrightarrow 0.$ So, using ($\star$), we have $\Delta F=\delta.$ Hence, we have a fundamental solution.
Case III: $n\ge 3$
Define functions $F$ on $\mathbb{R}^n$ as $F:x\mapsto\frac{|x|^{2-n}}{\omega_n(2-n)}$ and $F^t$ on $\mathbb{R}^n$ for all $t>0$ as $F^t:x\mapsto\frac{(|x|^2+t^2)^{(2-n)/2}}{\omega_n(2-n)}.$ Here, $\omega_n=\sigma(\mathcal{S}^{n-1})=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the volume of the unit sphere.
Now, $F$ is locally integrable function on $\mathbb{R}^n$ since for any compact $K\subset \mathbb{R}^n,$ there exists some $R>0$ such that $K\subset B(0,R)$ (by Heine Borel theorem), so using polar co-ordinates $$\int_K \left|\frac{|x|^{2-n}}{\omega_n(2-n)}\right|dx\le \int_{\mathcal{S}^{n-1}}\int_0^R\frac{|r|^{2-n}}{\omega_n(2-n)}r^{n-1}drd\sigma(x')$$$$=\frac{1}{\omega_n(2-n)}\cdot\frac{R^2}{2}\cdot \int_{\mathcal{S}^{n-1}}d\sigma(x')=\frac{R^2}{2(2-n)}<\infty$$
Now, $F^t$ is also locally integrable since $0\le F^t\le F.$ This is because $n>2.$
Now consider any $\phi\in\mathcal{D}(\mathbb{R}^n).$ Say $\text{supp}(\phi)\subset [-M/2,M/2]^n$ for some $M>0$ (since the support is compact and, hence, bounded by Heine Borel theorem). Now using the same steps as in Case I with the same justifications for usage of Fubini’s theorem and integration by parts, we have
$$\left\langle F^t,\frac{\partial^2}{\partial x_j^2}\phi\right\rangle=\int\left[-\frac{n}{\omega_n}(|x|^2+t^2)^{-n/2-1}x_j^2+\frac{1}{\omega_n}(|x|^2+t^2)^{-n/2}\right]\phi(x)dx$$
This holds for all $j\le n.$ On adding the above result for all $j\le n,$ we have $$\langle \Delta F^t, \phi\rangle=\langle F^t, \Delta \phi\rangle=\int \frac{nt^2}{\omega_n}(|x|^2+t^2)^{-n/2-1}\phi(x)dx\text{ }(\ddagger)$$
Define function $g$ on $\mathbb{R}^n$ as $g:x\mapsto\frac{n}{\omega_n}(|x|^2+1)^{-n/2-1}$ and $g^t$ on $\mathbb{R}^n$ for all $t>0$ as $g:x\mapsto t^{-n}g(t^{-1}x).$ Since $\phi\in\mathcal{D}(\mathbb{R}^n)$ is arbitrary, one can verify from $(\ddagger)$ that $\Delta F^t=g_t$ distributionally.
We note that $g\in L^1(\mathbb{R}^2)$ since (using polar coordinates and substituting $u=r^2/(r^2+1)$) $$\int |g|dm=\int gdm=\frac{n}{\omega_n}\int_{\mathcal{S}^{n-1}}\int_0^\infty(r^2+1)^{-n/2-1}r^{n-1}drd\sigma(x')=\frac{n\sigma(\mathcal{S}^{n-1})}{\omega_n}\int_0^1\frac{1}{2}u^{n/2-1}du=1<\infty$$
Using Proposition 9.1 (Folland), $\Delta F^t=g_t\longrightarrow\delta$ in $\mathcal{D}'(\mathbb{R}^n)$ as $t\longrightarrow 0.$ ($\star$)
Consider any sequence $(t_m)_{m=1}^\infty\subset (0,\infty)$ which converges to $0$ as $m$ approaches infinity. Now, for any $\phi\in\mathcal{D}(\mathbb{R}^n),$ we have $$\lim_{m\to\infty}\langle F^{t_m},\phi\rangle=\lim_{m\to\infty}\int \frac{(|x|^2+{t_m}^2)^{(2-n)/2}}{\omega_n(2-n)}\phi(z)dz=\int \lim_{m\to\infty}\frac{(|x|^2+{t_m}^2)^{(2-n)/2}}{\omega_n(2-n)}\phi(z)dz=\int \frac{|x|^{(2-n)/2}}{\omega_n(2-n)}\phi(z)dz=\langle F,\phi\rangle$$
This follows from the dominated convergence theorem since $F\in L^1_\text{loc}(\mathbb{R}^2)$ as shown before and $|F^{t_m}\phi|\le |F\phi|$ (where $|F\phi|\in L^1(\mathbb{R}^2)$ since $\phi$ has compact support and is bounded) for all $m\in\mathbb{N}.$
Since $\phi$ is arbitrary, we have $F^t\longrightarrow F$ (hence, also $\nabla F^t\longrightarrow \nabla F$) in $\mathcal{D}'(\mathbb{R}^2)$ as $t\longrightarrow 0.$ So, using ($\star$), we have $\Delta F=\delta.$ Hence, we have a fundamental solution.
